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3 years ago

How many rectangular faces does a hexagonal pyramid have ?

Mathematics

2 answers:

KiRa [710]3 years ago

4 0

The hexagonal pyramid Has 0 rectangular faces u can tell by the diagram below it has 6 triangle faces and 1 hexigon face I hope I helped if I did Mark me as brainliest

marusya05 [52]3 years ago

3 0

Zero should be the answer

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Exam Guidelines

Ganezh [65]

Answer:

it's A because it's not B or C

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2 years ago

(08.05)Some students reported how many textbooks they have. The dot plot shows the data collected:

steposvetlana [31]

The answer is B) There are exactly 4 students with 6 textbooks

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3 years ago

Tyler checked out two books from the school

kramer

Answer:

Step-by-step explanation:

224-146=78

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2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

What is the difference?

worty [1.4K]

Answer:

4th option

Step-by-step explanation:

Given

$\frac{x}{x^2+3x+2}$ - $\frac{1}{(x+2)(x+1)}$ ← expand denominator using FOIL

= $\frac{x}{x^2+3x+2}$ - $\frac{1}{x^2+3x+2}$

Since the denominators are common, then subtract the numerators

= $\frac{x-1}{x^2+3x+2}$

5 0

3 years ago